# Abstract Nonsense

## Generated Ideals and the Lattice of Ideals (Pt. I)

Point of Post: In this post we discuss the notion of generated ideals from subsets of a ring as well as the fact that the ideals of a ring form a complete modular lattice.

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Motivation

Much as was the case for normal subgroups of a group it’s true that there is some lattice like structure for ideals. In this post we shall describe the notion of  generated ideals and use it to prove that the poset $\left(\subseteq,\mathfrak{L}\left(R\right)\right)$ of ideals of $R$ is a complete modular lattice .

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Generated Ideals

Intuitively what we’d like to figure out is if we have some ring $R$ and some subset $A$ of $R$ can we find some left, right, or two-sided ideal which contains $A$ and is ‘as small as possible’ (i.e. minimal with respect to containment). The first step in doing this is the following observation:

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Theorem: Let $R$ be a ring and $\left\{\mathfrak{a}_\beta\right\}_{\beta\in\mathcal{B}}$ be a collection of ideals in $R$. Then, $\displaystyle \bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_{\beta}$ is an ideal of $R$.

Proof: We know that the intersection of subrings are subrings, and so it suffices to prove that $\displaystyle \bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta$ has the two-sided absorption property. To see this let $r\in R$ and $\displaystyle s\in\bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta$ be arbitrary. We note then that for every $\beta\in\mathcal{B}$ we have that $s\in\mathfrak{a}_\beta$ and so $rs\in\mathfrak{a}_\beta$. Thus, $rs\in\mathfrak{a}_\beta$ for all $\beta\in\mathcal{B}$ and so $\displaystyle rs\in\bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta$. Since $r$ was arbitrary we see that $\displaystyle \bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta$ has the left absorption property. A similar idea shows that it has the right absorption property and so $\displaystyle \bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta$ is an ideal as desired. $\blacksquare$

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A similar proof shows that in fact the arbitrary intersection of right or left ideals in a ring is also a left or right ideal respectively.

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With this in mind, if $R$ is a ring and $A\subseteq R$ we define the ideal generated by $A$, denoted $(A)$, to be equal to

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$\displaystyle \left(A\right)=\bigcap_{\mathfrak{a}\in\mathfrak{L}(R)\text{ s.t. }A\subseteq\mathfrak{a}}\mathfrak{a}\quad\bold{(1)}$

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Where $\mathfrak{L}(R)$ denotes the set of all ideals of $R$.  We define the notions of the generated left ideal and generated right ideal by replacing $\mathfrak{L}(R)$ with $\mathfrak{L}_L(R)$ and $\mathfrak{L}_R(R)$ respectively (where these denote the set of left and right ideals of $R$) and denote them $(A)_L$ and $(A)_R$. If $A=\{a_1,\cdots,a_n\}$ we denote $\left(A\right)$ as $(a_1,\cdots,a_n)$ (similarly for $(A)_L$ and $(A)_R$). An ideal $J$ (left, right, or two-sided) is called finitely generated

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So we have done what we’ve set out to do, namely show that any subset of a ring $R$ is contained inside some smallest ideal. That said, it isn’t very ‘nice looking’. As is the general case of math, once we’ve sucked the usefulness out of the abstract definition (in this case, it’s clear that our abstract desire of “minimality in terms of containment” is satisfied) we move on to a more practical definition, something we can actually work with (anyone having tried to do anything practical with singular homology understands this well). To do this though we need to make the concession that we are working with unital rings–but this really isn’t all too bad.

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Theorem: Let $R$ be a unital ring and $A\subseteq R$. Then, $(A)=RAR$, $(A)_L=RA$, and $(A)_R=AR$ where $RAR,RA$, and $AR$ are as previously defined.

Proof: We prove only that $(A)=RAR$ since the other cases are done similarly. To do this we first note that evidently $RAR\subseteq (A)$, since by definition every sum of the form $r_1a_1s_1+\cdots+r_na_ns_n$ where $r_k,s_k\in R$ and $a_k\in A$ is any ideal $\mathfrak{a}\supseteq A$. So, if we can show that $RAR\in\mathfrak{L}(R)$ we’ll be done since it will be an ideal containing $A$ (since $A=1_RA1_R\subseteq RAR$) and so $(A)\subseteq RAR$. So, to do this we let $r\in R$ and $r_1a_1s_1+\cdots+r_na_ns_n\in RAR$ be arbitrary, then $r(r_1a_1s_1+\cdots+r_na_ns_n)=(rr_1)a_1s_1+\cdots+(rr_n)a_ns_n\in RAR$ and $(r_1a_1s_1+\cdots+r_na_ns_n)r=r_1a_1(s_1r)+\cdots+r_na_n(s_nr)\in RAR$ and thus, by the arbitrariness of $r$ and $r_1a_1s_1+\cdots+r_na_ns_n$ we know that $RAR$ has the left and right absorption properties, thus it remains to show that $RAR\leqslant R$ (as abelian groups). To see this we merely note that if $r_1a_1s_1+\cdots+r_na_ns_n$ and $r'_1a'_1s'_1+\cdots+r'_ma'_ms'_m$ are in $RAR$ with $m\leqslant n$ then there difference is equal to $r''_1a''_1s''_1+\cdots+r''_{n+m}a''_{n+m}s''_{n+m}$ where $r''_k=r_k$ if $k\leqslant n$ and $r''_k=-r'_k$ if $k>n$, $a''_k=a_k$ if $k\leqslant n$ and $a''_k=-a'_k$ for $k>n$, and finally $s''_k=s_k$ for $k\leqslant n$ and $s''_k=-s'_k$ for $k>n$. But, this is clearly an element of $RAR$ and so $RAR\leqslant R$. The conclusion follows from previous discussion. $\blacksquare$

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Lattice of Ideals

With this notion of generated ideals in mind we can define the lattice structure on $\mathfrak{L}(R)$. Before we begin though we recall some of the basic concepts from order theory. Firstly, recall that if $(P,\leqslant)$ is a partially ordered set (poset) then a subset $S\subseteq P$ is said to have a supremum if there exist some $x\in P$ such that $s\leqslant x$ for all $s\in S$ and if $y\in P$ is such that $s\leqslant y$ for all $s\in S$ then $x\leqslant y$. It’s easy to prove that if $S$ has a supremum it’s unique, and so we may unambiguously denote it by $\sup S$ or $\bigvee S$ (or even $s_1\vee\cdots\vee s_n$ if $S=\{s_1,\cdots,s_n\}$). Similarly, $S$ is said to have an infimum if there exists some $x\in P$ such that $x\leqslant s$ for all $s\in S$ and if $y\in P$ is such that $s\leqslant y$ for all $s\in S$ then $y\leqslant x$. Once again, if $S$ has an infimum it’s unique and we may denote it $\inf S$ or $\bigwedge S$ (or even $s_1\wedge\cdots\wedge s_n$ if $S=\{s_1,\cdots,s_n\}$).

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A poset $(P,\leqslant)$ is called a complete lattice if $\sup S$ and $\inf S$ exist for all $S\subseteq P$. What we now claim is that given any ring $R$ one has that $\left(\mathfrak{L}(R),\subseteq\right)$ is a complete lattice. Indeed:

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Theorem: Let $R$ be a ring. Then $\left(\mathfrak{L}(R),\subseteq\right)$ is a complete lattice with $\displaystyle \sup \left\{\mathfrak{a}_\beta\right\}_{\beta\in\mathcal{B}}=\left(\bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta\right)$ and  $\displaystyle \inf\left\{\mathfrak{a}_\beta\right\}_{\beta\in\mathcal{B}}=\bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta$.

Proof: Let $\left\{\mathfrak{a}_\beta\right\}_{\beta\in\mathcal{B}}$ be any subset of $\mathfrak{L}(R)$. We have by previous theorem that $\displaystyle \bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta\in\mathfrak{L}(R)$, it’s clearly a lower bound for $\left\{\mathfrak{a}_\beta\right\}_{\beta\in\mathcal{B}}$ in $\left(P,\leqslant\right)$, and evidently if $E\in\mathfrak{L}(R)$ is such that $E\subseteq \mathfrak{a}_\beta$ for all $\beta\in\mathcal{B}$ then $\displaystyle E\subseteq\bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta$ and so $\displaystyle \bigcap_{\beta\in\mathcal{B}}\mathfrak{a}_\beta$ is the infimum for $\left\{\mathfrak{a}_\beta\right\}_{\beta\in\mathcal{B}}$. Similarly, we have by definition that $\displaystyle\left(\bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta\right)$ contains $\displaystyle \bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta$ and thus contains $\mathfrak{a}_\beta$ for each. Moreover, if $E\in\mathfrak{L}(R)$ contains $\mathfrak{a}_\beta$ for every $\beta\in\mathcal{B}$ then $E$ contains $\displaystyle \bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta$ and thus $\displaystyle \left(\bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta\right)\subseteq E$. Thus, $\displaystyle \bigcup_{\beta\in\mathcal{B}}\mathfrak{a}_\beta$ is the supremum of $\left\{\mathfrak{a}_\beta\right\}_{\beta\in\mathcal{B}}$. $\blacksquare$

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A similar statement holds for $\mathfrak{L}_L(R)$ and $\mathfrak{L}_R(R)$ with $(\cdot)_L$ and $(\cdot)_R$  in place of $(\cdot)$.

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

June 22, 2011 -

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